Question: What is the remainder when $2x^2-17x+47$ is divided by $x-5$?
Answer: \[
\begin{array}{c|cc c}
\multicolumn{2}{r}{2x} & -7 \\
\cline{2-4}
x-5 & 2x^2 &- 17x &+ 47  \\
\multicolumn{2}{r}{-2x^2} & +10x  \\ 
\cline{2-3}
\multicolumn{2}{r}{0} & -7x & +47   \\
\multicolumn{2}{r}{} &+ 7x  &-35   \\ 
\cline{3-4}
\multicolumn{2}{r}{} & 0& 12   \\

\end{array}
\]We cannot divide $12$ by $x-5$ since $12$ has lower degree. So the quotient is $2x-7$ and the remainder is $\boxed{12}$.